Let $f(x)=\cot(x)$. Find $f'\left(\dfrac{5\pi}{3}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\sqrt{3}$ (Choice B) B $\dfrac{1}{\sqrt{3}}$ (Choice C) C $-\dfrac{4}{3}$ (Choice D) D $\dfrac{4}{3}$
Answer: Let's first find $f'(x)$. Then, we can evaluate it at $x=\dfrac{5\pi}{3}$. Recall that the derivative of $\cot(x)$ is $-\dfrac{1}{\sin^2(x)}$, or $-\csc^2(x)$. [Is there a way to know this without memorizing?] So $f'(x)=-\dfrac{1}{\sin^2(x)}$. Now let's plug $x={\dfrac{5\pi}{3}}$ into $f'$ : $\begin{aligned} &\phantom{=}f'\left({\dfrac{5\pi}{3}}\right) \\\\ &=-\dfrac{1}{\sin^2\left({\dfrac{5\pi}{3}}\right)} \\\\ &=-\dfrac{1}{\left(-\dfrac{\sqrt 3}{2}\right)^2} \\\\ &=-\dfrac{1}{\left(\dfrac{3}{4}\right)} \\\\ &=-\dfrac{4}{3} \end{aligned}$ In conclusion, $f'\left(\dfrac{5\pi}{3}\right)=-\dfrac{4}{3}$.